Unformatted text preview: We can use Newton’s second
law to express the tension in each coupling bar, since friction is negligible: TA = ( m1 + m2 + m3 ) a
144424443 TB = ( m2 + m3 ) a
1442443 TC = m3a
1 24
43 Coupling bar A Coupling bar B Coupling bar C In these expressions a = 0.12 m/s2 remains constant. Consequently, the tension in a given
bar will change only if the total mass of the objects accelerated by that bar changes as a
result of the luggage transfer. Using ∆ (Greek capital delta) to denote a change in the usual
fashion, we can express the changes in the above tensions as follows:
∆TA = ∆ ( m1 + m2 + m3 ) a
4 14444 244444
3 ∆TB = ∆ ( m2 + m3 ) a
4 1444 24444
3 ∆TC = ( ∆m3 ) a
14 244
4
3 Coupling bar A Coupling bar B Coupling bar C 222 FORCES AND NEWTON'S LAWS OF MOTION SOLUTION
a. Moving luggage from car 2 to car 1 does not change the total mass m1 + m2 + m3, so
∆(m1 + m2 + m3) = 0 kg and ∆TA = 0 N . The transfer from car 2 to car 1 causes the total mass m2 + m3 to decrease by 39 kg, so
∆(m2 + m3) = –39 kg and ( ) ∆TB = ∆ ( m2 + m3 ) a = ( –39 kg ) 0.12 m/s2 = –4.7 N The transfer from car 2 to car 1 does not change the mass m3, so ∆m3 = 0 kg and
∆TC = 0 N .
b. Moving luggage from car 2 to car 3 does not change the total mass m1 + m2 + m3, so
∆(m1 + m2 + m3) = 0 kg and ∆TA = 0 N .
The transfer from car 2 to car 3 does not change the total mass m2 + m3, so ∆(m2 + m3) = 0
kg and ∆TB = 0 N .
The transfer from car 2 to car 3 causes the mass m3 to increase by 39 kg, so ∆m3 = +39 kg
and
∆TC = ( ∆m3 ) a = ( +39 kg ) 0.12 m/s 2 = +4.7 N ( ) ____________________________________________________________________________________________ 90. REASONING AND SOLUTION The distance required for the truck to stop is found from
x= 2
v 2 – v0 2a 2
( 0 m/s )2 – v0
= 2a The acceleration of the truck is needed. The frictional force decelerates the crate. The
maximum force that friction can supply is
fsMAX = µ sFN = µ smg
Newton's second law requires that
fsMAX = – ma so a = – µ sg
Now the stopping distance is Chapter 4 Problems x= 2
v0 2µs g ( 25 m/s )2
=
=
2 ( 0.650 ) ( 9.80 m/s 2 ) 223 49.1 m ____________________________________________________________________________________________ 91. REASONING AND SOLUTION
a. Newton's second law for block 1 (10.0 kg) is
T = m1a (1) Block 2 (3.00 kg) has two ropes attached each carrying a tension T. Also, block 2 only
travels half the distance that block 1 travels in the same amount of time so its acceleration is
only half of block 1's acceleration. Newton's second law for block 2 is then 2T − m2 g = − 1 m2 a
2 (2) Solving Equation (1) for a, substituting into Equation (2), and rearranging gives
T= 1+ 1m g
22
1 m /m
2
1
4 ( ) = 13.7 N b. Using this result in Equation (1) yields
a= T 13.7 N
=
= 1.37 m/s 2
m1 10.0 kg ____________________________________________________________________________________________ 92. REASONING AND SOLUTION
a. The force acting on the sphere which accelerates it is the horizontal component of the
tension in the string. Newton's second law for the horizontal motion of the sphere gives
T sin θ = ma
The vertical component of the tension in the string supports the weight of the sphere so
T cos θ = mg
Eliminating T from the above equations results in a = g tan θ . 224 FORCES AND NEWTON'S LAWS OF MOTION b. ( ) a = g tan θ = 9.80 m/s 2 tan 10.0° = 1.73 m/s 2 c. Rearranging the result of part a and setting a = 0 m/s2 gives θ = tan –1 ( a / g ) = 0°
____________________________________________________________________________________________ 93. SSM REASONING AND SOLUTION
a. The left mass (mass 1) has a tension T1 pulling it up. Newton's second law gives T1 – m1g = m1a (1) The right mass (mass 3) has a different tension, T3, trying to pull it up. Newton's second for
it is
(2)
T3 – m3g = – m3a
The middle mass (mass 2) has both tensions acting on it along with friction. Newton's
second law for its horizontal motion is
T3 – T1 – µ km2g = m2a (3) Solving Equation (1) and Equation (2) for T1 and T3, respectively, and substituting into
Equation (3) gives
( m − m1 − µk m2 ) g
a= 3
m1 + m2 + m3
Hence, ( ) 25.0 kg − 10.0 kg − ( 0.100 )( 80.0 kg ) 9.80 m/s2 a=
= 0.60 m/s 2
10.0 kg + 80.0 kg + 25.0 kg
b. From part a: (
)
T3 = m3(g − a) = ( 25.0 kg ) ( 9.80 m/s2 − 0.60 m/s 2 ) = 230 N
T1 = m1(g + a) = (10.0 kg ) 9.80 m/s 2 + 0.60 m/s2 = 104 N ____________________________________________________________________________________________ 94. REASONING AND SOLUTION
a. The static frictional force is responsible for accelerating the top block so that it does not
slip against the bottom one. The maximum force that can be supplied by friction is
fsMAX = µ sFN = µ sm1g Chapter 4 Problems 225 Newton's second law requires that fsMAX = m1a, so
a = µ sg
The force necessary to cause BOTH blocks to have this acceleration is
F = (m1 + m2)a = (m1 + m2)µ sg
F = (5.00 kg + 12.0 kg)(0.600)(9.80 m/s2) = 1.00 × 102 N
b. The maximum acceleration that the two block combination can have befo...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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